A 250 N child is in a swing that is attached to a 2.20 m long rope. Find the gra
ID: 1998095 • Letter: A
Question
A 250 N child is in a swing that is attached to a 2.20 m long rope. Find the gravitational potential energy of the child-Earth system relative to the child's lowest position for the following cases a. When the ropes are horizontal b. when the ropes make a 28.0 degree angle with the vertical c. Find the gravitational potential energy when the child is at the bottom of the circular are. A 15kg block is moving on a surface with a kinetic friction coefficient mu_k = 0.3, the initial velocity of the block is 1.5 m/s. What is the work done by the friction force by the time the speed ha reduced to 0.4m/s A 23kg block slides down an incline plane with an 18 degree angle and 3.2m in length. The kinetic friction coefficient is mu_k = 0.13. The block slides all the way down to the bottom of the plane a. Calculate the work done by the gravitational force b. Calculate the work done by the friction forces c. Calculate the velocity of the block when it gets to the bottom d. Compare the total potential energy when the block is at the top with the kinetic energy at the bottom, if different indicate what is the reason for the difference e. How much work was done on the block by conservative forces f. How much work was done by non-conservative forces on the block The curve shown in the Figure describes the changes in the potential energy of an object as a function of its position. By using graphical derivative, find the force as a function of time.Explanation / Answer
According to the given problem,
8.) GPE is always relative to some zero reference point. For swings, it makes sense to set GPE = 0 at the bottom of the swing's arc where theta = 0 deg relative to the vertical.
In general GPE = mgh; where mg = 350 N is the weight of the kid with mass m, g is the gravity field, and h is the height above the zero reference. In general, from ordinary trig, we have h = L(1 - cos(theta)) where L = 2.1 m is the length of each rope. So plug and chug into
a) GPE = mgh = 250*L(1 - cos(90)) = 250*2.2 = 550N.m You can do the math.
b) GPE = 250*2.2(1 - cos(28)) = 64.4N.m You can do the math.
c) GPE = 0 because we set it to zero or = 250*2.2(1 - cos(0)) = 0.
9.) Frinction force on the bloack is
f = ukmg = 0.3*15*9.81 = 44.145N
a = -f/m = -2.943m/s2
Using kinamatic equation for distance
v2-u2 = 2aS
S = 0.42 - 1.52/2*-2.943
S = 0.355 m
Work down by the friction is,
W = f *S
W = 14.675 J
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